ahmethungari Posted May 25, 2009 Posted May 25, 2009 (edited) Hi, Is there any solution for the following problem: [math]Ax = \lambda x + b[/math] Here [math]x[/math] seems to be an eigenvector of [math]A[/math] but with an extra translation vector [math]b[/math]. I cannot say whether [math]b[/math] is parallel to [math]x \quad[/math], ([math]b = cx[/math]). Thank you in advance for your help... Birkan Edited May 25, 2009 by ahmethungari
Bignose Posted May 25, 2009 Posted May 25, 2009 This forum doesn't support using $ as the math tags like regular LaTeX. You need to use (math) and (/math) but with square brackets in place of the beginning and ending $'s and the LaTeX will appear correctly.
Severian Posted May 25, 2009 Posted May 25, 2009 If [math]A- \lambda \mathbf{1}[/math] has an inverse, then the solution is [math]x= \left[ A- \lambda \mathbf{1} \right]^{-1} b[/math]
ahmethungari Posted May 26, 2009 Author Posted May 26, 2009 If [math]A- \lambda \mathbf{1}[/math] has an inverse, then the solution is [math]x= \left[ A- \lambda \mathbf{1} \right]^{-1} b[/math] The problem is we do not know the [math]\lambda[/math] either. Only [math]A[/math] and [math]b[/math] are knowns.
Severian Posted May 26, 2009 Posted May 26, 2009 Then you don't have enough information. If x is an N-dimensional vector, you have N+1 unknowns with only N equations. In a traditional eigenvalue problem, this doesn't matter because you don't need the magnitude of x (since every term is linear in x), but here the b term sets a scale.
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